A Scalar Equation of Motion

[motion.ar]

In classical mechanics, this topic presents a scalar equation of motion, which can be applied in any reference frame (rotating or non-rotating) (inertial or non-inertial) without the necessity of introducing fictitious forces.

If we consider two particles A and B of mass [latex]m_a[/latex] and [latex]m_b[/latex] respectively, then the scalar equation of motion, is given by:

[latex]\frac{1}{2}\,m_am_b\left[(\mathbf{v}_a-\mathbf{v}_b)^{2}+(\mathbf{a}_a-\mathbf{a}_b)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right]=\frac{1}{2}\,m_am_b\left[2\int\left(\frac{\mathbf{F}_a}{m_a}-\frac{\mathbf{F}_b}{m_b}\right){\cdot}\;d(\mathbf{r}_a-\mathbf{r}_b)+\left(\frac{\mathbf{F}_a}{m_a}-\frac{\mathbf{F}_b}{m_b}\right)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right][/latex]

where [latex]\mathbf{v}_a[/latex] and [latex]\mathbf{v}_b[/latex] are the velocities of particles A and B, [latex]\mathbf{a}_a[/latex] and [latex]\mathbf{a}_b[/latex] are the accelerations of particles A and B, [latex]\mathbf{r}_a[/latex] and [latex]\mathbf{r}_b[/latex] are the positions of particles A and B, and [latex]\mathbf{F}_a[/latex] and [latex]\mathbf{F}_b[/latex] are the net forces acting on particles A and B.

This scalar equation of motion is invariant under transformations between reference frames.

In addition, this scalar equation of motion would be valid even if Newton's three laws of motion were false.

Edited February 2, 2014 by motion.ar

[swansont]

It looks to me that if a and b are of equal mass, and are either co-located or subject to the same force, but have different speeds then the right side is zero but the left side is not.

[motion.ar]

Ok, swansont, but on the right side there is an indefinite integral. Therefore, in the previous equation, it follows that

 

[latex]\frac{1}{2}\,m_am_b\left[(\mathbf{v}_a-\mathbf{v}_b)^{2}+(\mathbf{a}_a-\mathbf{a}_b)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right]=constant[/latex]

Edited February 2, 2014 by motion.ar

[swansont]

Ok, swansont, but on the right side there is an indefinite integral. Therefore, in the previous equation, it follows that

 

[latex]\frac{1}{2}\,m_am_b\left[(\mathbf{v}_a-\mathbf{v}_b)^{2}+(\mathbf{a}_a-\mathbf{a}_b)\cdot(\mathbf{r}_a-\mathbf{r}_b)\right]=constant[/latex]

 

 

Of what value is that? You need to be able to evaluate the constant.

[motion.ar]

swansont, for example, if we consider a single particle A in a uniform gravitational field [latex]g[/latex] then the mechanical energy is constant.

 

[latex]\frac{1}{2} m_a \mathbf{v}_a^2 + m_a \, \mathbf{g} \cdot \mathbf{r}_a = constant[/latex]

 

This constant depends on the value of the mass and of the initial values of [latex]\mathbf{v}_a[/latex] and [latex]\mathbf{r}_a[/latex]

 

Similarly, the constant of the posted #3 depends on the values of the masses and of the initial values of [latex]\mathbf{v}_a, \mathbf{v}_b, \mathbf{a}_a, \mathbf{a}_b, \mathbf{r}_a, [/latex] and [latex]\mathbf{r}_b[/latex]

Edited February 2, 2014 by motion.ar

[swansont]

How do you determine the constant from the right-hand side of the equation, under the conditions I mentioned?

[motion.ar]

For example, through an imaginary experiment, this is, exposing an exercise.

In your imaginary experiment you should determine the initial conditions.

Edited February 2, 2014 by motion.ar

[studiot]

 

If we consider two particles A and B of mass and respectively, then the scalar equation of motion, is given by

 

Perhaps I'm being thick tonight but I fail to see why B has any effect on the motion of A, within the conditions you have stated.

[motion.ar]

In the equation of the topic # 1, a particle A is related with another particle B.

 

In the above example, it is possible that a particle D exerts a force [latex]\mathbf{F}_a[/latex] on a particle A, and a particle E exerts a force [latex]\mathbf{F}_b[/latex] on a particle B, where [latex]\mathbf{F}_a=\mathbf{F}_b[/latex] and [latex]m_a=m_b[/latex]

 

In the above example, it is possible that the particle A does not exert any force on the particle B, and the particle B does not exert any force on particle A.

Edited February 3, 2014 by motion.ar

[studiot]

 

In the equation of the topic # 1, a particle A is related with another particle B.

 

In the above example, it is possible that a particle D exerts a force on a particle A, and a particle E exerts a force on a particle B, where and

 

In the above example, it is possible that the particle A does not exert any force on the particle B, and the particle B does not exert any force on particle A.

 

 

Thank you for adding information to your original statement.

 

But that does not really answer my question.

 

You must have some origin O to reference you forces Fa and Fb to as well as your postion vectors.

Although you have not explicitly stated this I assume they all have a common origin.

 

So you then appear to create an expresion for the relative motion between some particle A and another B, the only connection being the vector origin you have chosen.

 

If we wish to determine the motion of particle B from you equation, What, why and how do we choose particle A, given that A had no influence on the motion of B?

[motion.ar]

studiot, the reference point is the origin O of the reference frame which makes measurements of: [latex]\mathbf{r_{a_o}}, \mathbf{v_{a_o}}, \mathbf{a_{a_o}}, \mathbf{r_{b_o}}, \mathbf{v_{b_o}}, \mathbf{a_{b_o}}, [/latex] etc.

Edited February 4, 2014 by motion.ar

[studiot]

 

But that does not really answer my question.

 

So I will repeat my question yet again.

 

Object A can move quite independently of object B.

 

So how does the relative velocity help?

or if you like what is the gain in introducing object A to monitor the motion of object B against?

Edited February 4, 2014 by studiot